关键路径和最短路径详解过程

1、求下图从事件0出发的关键路径，要求详细过程。

1、事件Vj 可能的最早发生时间ve(j) Ve(0)=0;

Ve(1)=ve(0)+weight()=0+5=5; Ve(2)=ve(0)+weight()=0+9=9; Ve(3)=ve(0)+weight()=0+14=14; Ve(4)=ve(1)+weight()=5+4=9;

Ve(5)=max{ve(2)+weight(),ve(4)+weight()}=max{9+10,9+6}=19; Ve(6)=ve(3)+weight()=14+3=17;

Ve(7)=max{ve(3)+weight(),ve(6)+weight()}=max{14+7,17+5}=22; Ve(8)=max{ve(6)+weight(),ve(7)+weight()}=max{17+5,22+8}=30; Ve(9)=max{ve(4)+weight(),ve(5)+weight(),ve(8)+weight()}= Max{9+12,19+10,30+18}=48;

2、事件vi可能的最晚发生时间vl(i)

Vl(9)=48;

Vl(8)=vl(9)-weight()=48-18=30;

Vl(7)=vl(8)-weight()=30-8=22;

Vl(6)=min{ve(7)-weight(),ve(8)-weight()}=min{22-5,30-5}=min{17,25}=17; Vl(5)=vl(9)-weight()=48-10=38;

Vl(4)=min{vl(5)-weight(),vl(9)-weight()}=min{38-6,48-12}=min{32,36}=32; Vl(3)=min{vl(6)-weight(),vl(7)-weight()}=min{17-3,22-7}=min{14,15}=14 Vl(2)=vl(5)-weight()=38-10=28;

Vl(1)=vl(4)-weight()=32-4=28;

Vl(0)=min{vl(1)-weight(),vl(2)-weight(),vl(3)-weight()}= Min{28-5,28-9,14-14}=min{23,19,0}=0;

3、活动a(k)=的最早开始时间E(k)

E(0)=ve(0)=0

E(1)=ve(0)=0

E(2)=ve(0)=0

E(3)=ve(1)=5

E(4)=ve(2)=9

E(5)=ve(3)=14

E(6)=ve(3)=14

E(7)=ve(4)=9

E(8)=ve(6)=17

E(9)=ve(4)=9

E(10)=ve(5)=19

E(11)=ve(6)=17

E(12)=ve(7)=22

E(13)=ve(8)=30

4、活动a(k)的最晚开始时间L(k)

L(0)=vl(1)-weight()==28-5=23

L(1)=vl(2)-weight()==28-9=21

L(2)=vl(3)-weight()==14-14=0

L(3)=vl(4)-weight()==32-4=28

L(4)=vl(5)-weight()==38-10=28

L(5)=vl(6)-weight()==17-3=14

L(6)=vl(7)-weight()==22-7=15

L(7)=vl(5)-weight()==38-6=32

L(8)=vl(7)-weight()==22-5=17

L(9)=vl(9)-weight()==48-12=36

L(10)=vl9)-weight()==48-10=38

L(11)=vl(8)-weight()==30-5=34

L(12)=vl(8)-weight()==30-8=22

L(13)=vl(9)-weight()==48-18=30

L(0)-E(0)=23-0=23

L(1)-E(1)=21-0=21

L(2)-E(2)=0-0=0

L(3)-E(3)=28-5=23

L(4)-E(4)=28-9=19

L(5)-E(5)=14-14=0

L(6)-E(6)=15-14=1

L(7)-E(7)=32-9=23

L(8)-E(8)=17-17=0

L(9)-E(9)=36-9=27

L(10)-E(10)=38-19=19

L(11)-E(11)=34-17=17

L(12)-E(12)=22-22=0

L(13)-E(13)=30-30=0

5、题中图的关键路径如下图所示：

2、对于下图所示的有向图，试利用Dijkstra算法求源点1到其他各顶点的最短路径，可参考教材P207图7.29中的表格形式给出计算过程。